It is well known that in any triangle $ABC$, the lines $AH$ and $AO$ are reflections of each other across the angle bisector of $A$ (points $H$ and $O$ are isogonal conjugates). Take an arbitrary triangle $ABD$ and it follows that the line $AC$ is just an extension of the line $AH$. Hence it's reflection is the line $AO$, which obviously contains point O.

(Solution copied from this thread https://artofproblemsolving.com/community/u404977h1657134p10507651 under request of admin) Let $F$ be the second intersection of line $AO$ with the circumcircle. $\angle FAD = \frac{\widehat{FD}}{2} = \frac{\widehat{FA}-\widehat{AD}}{2}= 90 - \angle B = \angle CAB$. Thus the line $AO$ is the reflection of $AC$ across the bisector, as it is isogonal to $AC$, and we are done. This really is a well-known result, it's essentially just showing that the altitude and the circumradius from $A$ are isogonal.